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This hints that $E(|S_n|)\,\!$, the expected translation distance after ''n'' steps, should be of the order of $\sqrt n$. In fact,

$$\lim_{n\to\infty} \frac{E(|S_n|)}{\sqrt n}= \sqrt{\frac 2{\pi}}.$$ (from http://en.wikipedia.org/wiki/Random_walk#One-dimensional_random_walk)

Why is it like this? After looking at Wikipedia contents all below, I was not able to find proof - so can anyone provide me proof?

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Random work is kind of poetry. –  Did Dec 14 '12 at 7:06
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up vote 1 down vote accepted

In addition to the arguments in the link in the comment above, there is this possibility to use a relationship between central limit theorems and moment convergence.

If $S_n$ is a sum of $n$ i.i.d. random variables satisfying $E X=0$, $E X^2=1$ and more generally $E|X|^r < \infty$ for some $r\geq 2$, then $E|S_n/ \sqrt{n}|^r$ converges to $E |Z|^r $ where $Z$ is a standard normal. (See theorem 5.1 on page 353 in Gut's "Probabity: a graduate course")

Applying this to $E | S_n/ \sqrt{n}|$ yields $\sqrt{2/\pi}$ because this the expectation of $E |Z| $ for a standard normal.

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